【c语言】推箱子

yyqzjw2023-10-03 13:58

所需知识：c语言枚举，数组，for循环，while循环，switch,case语句，图形库相关函数

1.调整控制台窗口大小

c 复制代码 
#define  _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>

#include <stdio.h>
int main()
{


	system("mode con lines=15 cols=25");//调整窗口大小
	return 0;
}

2.清掉控制台屏幕上的字

c 复制代码 
#define  _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>

#include <stdio.h>
int main()
{


	system("mode con lines=15 cols=25");
	system("cls");//清屏操作
	getchar();//不让程序退出，等待读字符
	return 0;
}

3. 枚举类型定义地图中空地，墙，目的地，箱子，玩家

c 复制代码 
enum  Mine
{
	SPACE,  //空地
	WALL,//墙
	DEST,  //目的地
	BOX,  //箱子
	PLAYER//玩家
};

4.定义二维数组做地图，并且打印出来看看效果

c 复制代码 
//定义一个二维数组，做地图 空地0  墙1  目的地2  箱子3  玩家4    箱子在目的地 5  玩家在目的地6,与枚举类型对应上了
int map[10][10] =
{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,1,1,1,0,0,0,0},
	{0,0,0,1,2,1,1,1,1,0},
	{0,1,1,1,3,0,3,2,1,0},
	{0,1,2,3,4,0,1,1,1,0},
	{0,1,1,1,1,3,1,0,0,0},
	{0,0,0,0,1,2,1,0,0,0},
	{0,0,0,0,1,1,1,0,0,0},
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
};

void printmap()
{for(int i=0;i<10;i++)
{
	for (int j = 0; j < 10; j++)
	{
		printf("%d ", map[i][j]);





	}
	printf("\n");




}






}

将printmap()在main中调用

为了让程序不会输入字符后退出，加上while循环

c 复制代码 
int main()
{

	while (1)
	{
		system("mode con lines=15 cols=25");
		system("cls");//清屏操作
		printmap();
		getchar();//不让程序退出，等待读字符
	}
	return 0;
}

5.修改printmap函数为gamedraw()函数

为了保证游戏的美观性，我们将对应的数字转化为好看的图案

使用两层循环遍历二维数组，在使用switch已经将对应的数字用图案代替，此时我们要下载搜狗输入法

c 复制代码 
void gamedraw()
{
	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			 
				switch (map[i][j])
				{case SPACE://如果二维数组元素为0
					printf("  ");  //空地        //一个中文字符相当于二个英文字符
					break;
				case WALL://如果二维数组元素为1
					printf("■");//墙
					break;
				case DEST://如果二维数组元素为2
					printf("☆");//目的地
					break;
				case BOX://如果二维数组元素为3
					printf("□");//箱子
					break;
				case PLAYER://如果二维数组元素为4
					printf("♀");//玩家
					break;
				case PLAYER+DEST://如果二维数组元素为6
					printf("♂");//玩家在目的地
					break;
				case BOX+DEST://如果二维数组元素为5
					printf("★");箱子在目的地
					break;
                }






		}
				
printf("\n");
	}





}

对应数字和枚举变量类型对应上，PLAYER+DEST表示玩家如果出现在目的地的话对应数字为6，BOX+DEST表示箱子在目的地，将主函数中的printmap用gamedraw()换掉。

6.按键控制移动

要想使玩家移动，就先得确定玩家的坐标，通过遍历二维数组，找到数组中数字等于4的和4+2的，后者表示玩家在目的地，也要获取坐标，当找到玩家坐标时，要跳出循环，而break只能跳出一次循环，所以使用goto函数，可以直接跳出多层循环.

c 复制代码 
int i = 0; int j = 0;//定义不要在for循环里面，要不然出作用域就会被销毁
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (map[i][j] == PLAYER||map[i][j] == PLAYER+DEST)
			{
				goto end;
			}
		}
	}
end:;//找到直接来这里

在使用_getch()函数将按键的信息放入ch字符变量中，如果不知道上下左右对应的键值，我们可以打印出ch看看，我们把这些都封装成一个函数 keyevent()获取玩家坐标，并且读取键盘按下的信息

c 复制代码 
void keyevent()
{
	int i = 0; int j = 0;
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (map[i][j] == PLAYER || map[i][j] == PLAYER + DEST)
			{
				goto end;
			}
		}
	}
end:;
	char ch = _getch();
	printf("%d    %c", ch, ch);//w 119 a 97 s 115  d 100

















}

通过printf("%d %c", ch, ch);

得到虚拟键值为//w 119 a 97 s 115 d 100，然后注释掉printf("%d %c", ch, ch);然后通过swich,case 语句分别处理上下左右按键按下后的处理，

c 复制代码 
	switch (ch)
	{
	case 119:
	case 'w ':
	case 'W':
		break;
	case 97:
	case 'a ':
	case'A':
		break;
	case 115:
	case 's ':
	case'S':
		break;
	case 100:
	case 'd ':
	case'D':
		break;
	}

比如说按下w键，如果玩家上面是空地或者是目的地的话，玩家可以直接挪过去，因为没有障碍物阻碍，如果玩家的坐标为map[i][j];则玩家上面的的坐标就是map[i-1][j];要做的是

c 复制代码 
 if (map[i - 1][j] == SPACE||map[i - 1][j] == DEST)
		 {
			 map[i - 1][j] += PLAYER;
			 map[i][j] -= PLAYER;



		 }

如果map[i][j]只有玩家的话，上移动玩家就会map[i][j]=0，该位置变为空地，如果map[i][j]是玩家加目的地，上移动玩家就会map[i][j]=DEST，变为单纯的目的地.

如果玩家上面一个位置是箱子或者是箱子加目的地，就要看玩家上面的上面是什么了，如果是空地，或者是目的地，就可以推动，map[i - 2][j]是玩家上面的上面的坐标要做的是

c 复制代码 
 else  if(map[i-1][j]==BOX||map[i-1][j]==BOX+DEST)
		 {
			 if (map[i - 2][j] == SPACE || map[i - 2][j] == DEST)
			 {//完成玩家上面有箱子，箱子的上面是空地或者是目的地都可以推动
				 map[i - 2][j] += BOX;//玩家上面的上面加一个箱子
				 map[i - 1][j] = map[i - 1][j] - BOX + PLAYER;//玩家的上面减去一个箱子加上一个玩家
				 map[i][j] -= PLAYER;//玩家消失在原来位置





			 }



		 }

处理完的函数上键

c 复制代码 
case 119:
	case 'w ':
	case 'W':
		if (map[i - 1][j] == SPACE || map[i - 1][j] == DEST)
		{
			map[i - 1][j] += 4;
			map[i][j] -= 4;



		}
		else  if (map[i - 1][j] == BOX || map[i - 1][j] == BOX + DEST)
		{
			if (map[i - 2][j] == SPACE || map[i - 2][j] == DEST)
			{
				map[i - 2][j] += BOX;
				map[i - 1][j] = map[i - 1][j] - BOX + PLAYER;
				map[i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;

向上推了一下，如果懂了上键怎么移动，别的也就会处理了

整体的keyevent()

c 复制代码 
void keyevent()
{
	int i = 0; int j = 0;
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (map[i][j] == PLAYER || map[i][j] == PLAYER + DEST)
			{
				goto end;
			}
		}
	}
end:;
	char ch = _getch();
	switch (ch)
	{
	case 119:
	case 'w ':
	case 'W':
		if (map[i - 1][j] == SPACE || map[i - 1][j] == DEST)
		{
			map[i - 1][j] += PLAYER;
			map[i][j] -=  PLAYER;



		}
		else  if (map[i - 1][j] == BOX || map[i - 1][j] == BOX + DEST)
		{
			if (map[i - 2][j] == SPACE || map[i - 2][j] == DEST)
			{
				map[i - 2][j] += BOX;
				map[i - 1][j] = map[i - 1][j] - BOX + PLAYER;
				map[i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 97:
	case 'a ':
	case'A':
		if (map[i][j - 1] == SPACE || map[i][j - 1] == DEST)
		{
			map[i][j - 1] += PLAYER;
			map[i][j] -=  PLAYER;



		}
		else  if (map[i][j - 1] == BOX || map[i][j - 1] == BOX + DEST)
		{
			if (map[i][j - 2] == SPACE || map[i][j - 2] == DEST)
			{
				map[i][j - 2] += BOX;
				map[i][j - 1] = map[i][j - 1] - BOX + PLAYER;
				map[i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 115:
	case 's ':
	case'S':
		if (map[i + 1][j] == SPACE || map[i + 1][j] == DEST)
		{
			map[i + 1][j] += PLAYER;
			map[i][j] -= PLAYER;



		}
		else  if (map[i + 1][j] == BOX || map[i + 1][j] == BOX + DEST)
		{
			if (map[i + 2][j] == SPACE || map[i + 2][j] == DEST)
			{
				map[i + 2][j] += BOX;
				map[i + 1][j] = map[i + 1][j] - BOX + PLAYER;
				map[i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 100:
	case 'd ':
	case'D':
		if (map[i][j + 1] == SPACE || map[i][j + 1] == DEST)
		{
			map[i][j + 1] +=  PLAYER;
			map[i][j] -=  PLAYER;



		}
		else  if (map[i][j + 1] == BOX || map[i][j + 1] == BOX + DEST)
		{
			if (map[i][j + 2] == SPACE || map[i][j + 2] == DEST)
			{
				map[i][j + 2] += BOX;
				map[i][j + 1] = map[i][j + 1] - BOX + PLAYER;
				map[i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	}


}

7.多组地图的制作，修改map数组

定义全局变量level=0;表示关数，

将二维数组改为三维数组，三维数组的每一个元素就是二维数组，就是一个地图.

c 复制代码 
int map[3][10][10] =
{
	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,1,1,1,0,0,0,0},
	{0,0,0,1,2,1,1,1,1,0},
	{0,1,1,1,3,0,3,2,1,0},
	{0,1,2,3,4,0,1,1,1,0},
	{0,1,1,1,1,3,1,0,0,0},
	{0,0,0,0,1,2,1,0,0,0},
	{0,0,0,0,1,1,1,0,0,0},
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},

	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,1,1,0,0,1,1,0,0},
	{0,1,0,2,1,1,2,0,1,0},
	{1,0,0,0,3,0,0,0,0,1},
	{1,0,0,0,4,3,0,0,0,1},
	{0,1,0,0,3,3,0,0,1,0},
	{0,0,1,0,0,0,0,1,0,0},
	{0,0,0,1,2,2,1,0,0,0},
	{0,0,0,0,1,1,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},
	{
	{0,0,0,0,1,0,0,0,0,0},
	{0,0,0,1,0,1,0,0,0,0},
	{0,0,1,2,3,0,1,0,0,0},
	{0,1,0,0,0,0,0,1,0,0},
	{1,2,3,0,4,0,0,0,1,0},
	{0,1,0,0,0,0,0,3,2,1},
	{0,0,1,0,3,0,0,0,1,0},
	{0,0,0,1,2,0,0,1,0,0},
	{0,0,0,0,1,0,1,0,0,0},
	{0,0,0,0,0,1,0,0,0,0}
	}
};

将所有map[][]改成map[level][][];每通过一关，level++;

8.通关判断

循环遍历二维数组，如果有map[level][i][j]==BOX，说明有箱子没有推到目的地，返回false ，循环结束如果没有的话，说明通过此关，返回true,该判断函数返回布尔类型的值

c 复制代码 
bool jude()
{

	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			if (map[level][i][j] == BOX)
			{
				return false;
			}
		}
	}






}
c 复制代码 
	if (jude())
		{
			level++;
			if (level > 2)
			{
				
				printf("oioioioioioioioi奥哈呦学妹你通过了！");
			
				_getch();
				break;
			}
		}

主函数添加如果jude()返回1，然后就是通关，然后level++;

变成三维数组中的第二个元素，也就是换了地图，当level>2,表示通关，因为只设置了三个图

9.程序源码（无图形库）

c 复制代码 
#define  _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <conio.h>
#include <stdio.h>
enum  Mine
{
	SPACE,  //空地
	WALL,//墙
	DEST,  //目的地
	BOX,  //箱子
	PLAYER//玩家





};
int level = 0;
//定义一个二维数组，做地图 空地0  墙1  目的地2  箱子3  玩家4    箱子在目的地 5  玩家在目的地6,与枚举类型对应上了
int map[3][10][10] =
{
	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,1,1,1,0,0,0,0},
	{0,0,0,1,2,1,1,1,1,0},
	{0,1,1,1,3,0,3,2,1,0},
	{0,1,2,3,4,0,1,1,1,0},
	{0,1,1,1,1,3,1,0,0,0},
	{0,0,0,0,1,2,1,0,0,0},
	{0,0,0,0,1,1,1,0,0,0},
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},

	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,1,1,0,0,1,1,0,0},
	{0,1,0,2,1,1,2,0,1,0},
	{1,0,0,0,3,0,0,0,0,1},
	{1,0,0,0,4,3,0,0,0,1},
	{0,1,0,0,3,3,0,0,1,0},
	{0,0,1,0,0,0,0,1,0,0},
	{0,0,0,1,2,2,1,0,0,0},
	{0,0,0,0,1,1,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},
	{
	{0,0,0,0,1,0,0,0,0,0},
	{0,0,0,1,0,1,0,0,0,0},
	{0,0,1,2,3,0,1,0,0,0},
	{0,1,0,0,0,0,0,1,0,0},
	{1,2,3,0,4,0,0,0,1,0},
	{0,1,0,0,0,0,0,3,2,1},
	{0,0,1,0,3,0,0,0,1,0},
	{0,0,0,1,2,0,0,1,0,0},
	{0,0,0,0,1,0,1,0,0,0},
	{0,0,0,0,0,1,0,0,0,0}
	}
};

void gamedraw()
{
	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			 
				switch (map[level][i][j])
				{case SPACE:
					printf("  ");          //一个中文字符相当于二个英文字符
					break;
				case WALL:
					printf("■");
					break;
				case DEST:
					printf("☆");
					break;
				case BOX:
					printf("□");
					break;
				case PLAYER:
					printf("♀");
					break;
				case PLAYER+DEST:
					printf("♂");
					break;
				case BOX+DEST:
					printf("★");
					break;
                }






		}
				
printf("\n");
	}





}
void keyevent()
{
	int i = 0; int j = 0;
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (map[level][i][j] == PLAYER || map[level][i][j] == PLAYER + DEST)
			{
				goto end;
			}
		}
	}
end:;
	char ch = _getch();
	switch (ch)
	{
	case 119:
	case 'w ':
	case 'W':
		if (map[level][i - 1][j] == SPACE || map[level][i - 1][j] == DEST)
		{
			map[level][i - 1][j] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i - 1][j] == BOX || map[level][i - 1][j] == BOX + DEST)
		{
			if (map[level][i - 2][j] == SPACE || map[level][i - 2][j] == DEST)
			{
				map[level][i - 2][j] += BOX;
				map[level][i - 1][j] = map[level][i - 1][j] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 97:
	case 'a ':
	case'A':
		if (map[level][i][j - 1] == SPACE || map[level][i][j - 1] == DEST)
		{
			map[level][i][j - 1] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i][j - 1] == BOX || map[level][i][j - 1] == BOX + DEST)
		{
			if (map[level][i][j - 2] == SPACE || map[level][i][j - 2] == DEST)
			{
				map[level][i][j - 2] += BOX;
				map[level][i][j - 1] = map[level][i][j - 1] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 115:
	case 's ':
	case'S':
		if (map[level][i + 1][j] == SPACE || map[level][i + 1][j] == DEST)
		{
			map[level][i + 1][j] += PLAYER;
			map[level][i][j] -= PLAYER;



		}
		else  if (map[level][i + 1][j] == BOX || map[level][i + 1][j] == BOX + DEST)
		{
			if (map[level][i + 2][j] == SPACE || map[level][i + 2][j] == DEST)
			{
				map[level][i + 2][j] += BOX;
				map[level][i + 1][j] = map[level][i + 1][j] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 100:
	case 'd ':
	case'D':
		if (map[level][i][j + 1] == SPACE || map[level][i][j + 1] == DEST)
		{
			map[level][i][j + 1] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i][j + 1] == BOX || map[level][i][j + 1] == BOX + DEST)
		{
			if (map[level][i][j + 2] == SPACE || map[level][i][j + 2] == DEST)
			{
				map[level][i][j + 2] += BOX;
				map[level][i][j + 1] = map[level][i][j + 1] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	}


}
bool jude()
{

	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			if (map[level][i][j] == BOX)
			{
				return false;
			}
		}
	}






}








int main()
{
	system("mode con lines=15 cols=25");
	//system("cls");//清屏操作
	while (1)
	{
		
	
		
		gamedraw();
		
		//_getch();
		if (jude())
		{
			level++;
			if (level > 2)
			{
				
				printf("oioioioioioioioi奥哈呦学妹你通过了！");
			
				_getch();
				break;
			}
		}keyevent();
		
	}
	getchar();//不让程序退出，等待读字符
	return 0;
}

10.演示1

20231002_124830

11.加图形库版本

12.头文件增加

c 复制代码 
#include <graphics.h >

13.定义保存空地，目的地，玩家，箱子，墙，箱子推到目的地图片的类

c 复制代码 
IMAGE  ima_all[6];

14.将图片image文件放在.cpp文件同目录下

image文件夹是自己创建的，用于放推箱子的素材，就是图片，图片可以在网上自己找推箱子的图片

15.加载图片函数

c 复制代码 
void loadimg()
{
	
	for (int i = 0; i < 6; i++)
	{
		char file[20] = "";
		sprintf(file,"./images/%d.png", i);
		loadimage(ima_all + i,file, 40, 40);
		

	}





}

为什么这么加载图片，这里我们将照片命名为了0，1,2，3,4,5，要将这六个照片都加载进去，六个照片的相对路径里面只有照片名字不一样，我们可以循环将每个照片对应的相对路径的字符串放入到file字符串数组file中去，使用sprintf,如果这里不知道sprintf的用法，可以去看看我文件操作那一篇，文件操作，使用loadimage函数将六个图片加载进去，loadimage第一个参数是图片的首地址，第二个参数是该图片的相对路径，第三个，第四个参数是分辨率，也就是大小，后两个可以在

可以看到4241，这里统一用4040；

16. gamedraw函数的修改

因为使用图形库的话就不用之前的gamedraw函数打印图案，而是将图片贴上去

c 复制代码 
void gamedraw()
{
	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			int x = j * 40;
			int y = i * 40;

			switch (map[level][i][j])
			{
			case SPACE:
				putimage(x, y, ima_all+2);         //一个中文字符相当于二个英文字符
				break;
			case WALL:
				putimage(x, y, ima_all+1);
				break;
			case DEST:
				putimage(x, y, ima_all+4);
				break;
			case BOX:
				putimage(x, y, ima_all+3);
				break;
			case PLAYER:
				putimage(x, y, ima_all);
				break;
			case PLAYER + DEST:
				putimage(x, y, ima_all);
				break;
			case BOX + DEST://就是箱子推到目的地
				putimage(x, y, ima_all+5);
				break;
			}






		}

		
	}





}

由上图可知i,j和x,y反了过来，使用putimage函数将每个图片贴上去，putimage前两个参数为图片要贴在界面上左上角的坐标，第三个参数是要贴图片对应的地址，之前命名的照片在数组ima_all[6];是对应上的，就是说ima_all+3地址对应的就是对应的这个图片

17.主函数修改

c 复制代码 
int main()
{
	initgraph(400, 400);
	loadimg();
	//system("mode con lines=15 cols=25");
	//system("cls");//清屏操作
	while (1)
	{
		
	
		
		gamedraw();
		
		//_getch();
		if (jude())
		{
			level++;
			if (level > 2)
			{
				
				//printf("oioioioioioioioi奥哈呦学妹你通过了！");
			
				_getch();
				break;
			}
		}keyevent();
		
	}
	getchar();//不让程序退出，等待读字符
	return 0;
}

不使用控制台显示地图，初始化界面，在界面上显示地图，由于一个图片是40乘40的，二维数组是10乘10的，所以界面行列都应该是400，所以initgraph(400, 400);由于不用控制台，所以也就不使用printf函数，也不用将控制台大小改变，也不要清屏.

当你此时开始编译运行的时候会出现下列错误

解决方案：调试-》属性-》高级-》字符集-》多字符集

18.游戏通关显示

如果level>2

c 复制代码 
settextcolor(BLACK);//字体颜色
				settextstyle(25, 0, "微软雅黑");//字体风格
				setbkmode(TRANSPARENT);//字体背景透明
				
				outtextxy(100, 100, "oioioioioioioioi奥哈呦学妹你通过了！");//字体显示位置，以及内容
				_getch();
				break;

19.程序源码（带图形库版）

c 复制代码 
#define  _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>//_getch()函数头文件
#include <stdbool.h>//bool类型的头函数
#include <graphics.h >//图形库头文件
//定义一个二维数组，做地图
//空地0  墙1  目的地2  箱子3  玩家4    箱子在目的地 5  玩家在目的地6
IMAGE  ima_all[6];
int level = 0;
void loadimg()
{
	
	for (int i = 0; i < 6; i++)
	{
		char file[20] = "";
		sprintf(file,"./images/%d.png", i);
		loadimage(ima_all + i,file, 40, 40);
		

	}





}
enum  Mine
{
	SPACE,  //空地
	WALL,//墙
	DEST,  //目的地
	BOX,  //箱子
	PLAYER//玩家





};
int map[3][10][10] =
{
	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,1,1,1,0,0,0,0},
	{0,0,0,1,2,1,1,1,1,0},
	{0,1,1,1,3,0,3,2,1,0},
	{0,1,2,3,4,0,1,1,1,0},
	{0,1,1,1,1,3,1,0,0,0},
	{0,0,0,0,1,2,1,0,0,0},
	{0,0,0,0,1,1,1,0,0,0},
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},

	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,1,1,0,0,1,1,0,0},
	{0,1,0,2,1,1,2,0,1,0},
	{1,0,0,0,3,0,0,0,0,1},
	{1,0,0,0,4,3,0,0,0,1},
	{0,1,0,0,3,3,0,0,1,0},
	{0,0,1,0,0,0,0,1,0,0},
	{0,0,0,1,2,2,1,0,0,0},
	{0,0,0,0,1,1,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},
	{
	{0,0,0,0,1,0,0,0,0,0},
	{0,0,0,1,0,1,0,0,0,0},
	{0,0,1,2,3,0,1,0,0,0},
	{0,1,0,0,0,0,0,1,0,0},
	{1,2,3,0,4,0,0,0,1,0},
	{0,1,0,0,0,0,0,3,2,1},
	{0,0,1,0,3,0,0,0,1,0},
	{0,0,0,1,2,0,0,1,0,0},
	{0,0,0,0,1,0,1,0,0,0},
	{0,0,0,0,0,1,0,0,0,0}
	}
};



void gamedraw()
{
	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			int x = j * 40;
			int y = i * 40;

			switch (map[level][i][j])
			{
			case SPACE:
				putimage(x, y, ima_all+2);         //一个中文字符相当于二个英文字符
				break;
			case WALL:
				putimage(x, y, ima_all+1);
				break;
			case DEST:
				putimage(x, y, ima_all+4);
				break;
			case BOX:
				putimage(x, y, ima_all+3);
				break;
			case PLAYER:
				putimage(x, y, ima_all);
				break;
			case PLAYER + DEST:
				putimage(x, y, ima_all);
				break;
			case BOX + DEST:
				putimage(x, y, ima_all+5);
				break;
			}






		}

		
	}





}
void keyevent()
{
	int i = 0; int j = 0;
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (map[level][i][j] == PLAYER||map[level][i][j] == PLAYER+DEST)
			{
				goto end;
			}
		}
	}
end:;
	char ch = _getch();
	//printf("%d    %c", ch, ch);//w 119 a 97 s 115  d 100
	switch (ch)
	{case 119:
	 case 'w ':
	 case 'W':
		 if (map[level][i - 1][j] == SPACE||map[level][i - 1][j] == DEST)
		 {
			 map[level][i - 1][j] += 4;
			 map[level][i][j] -= 4;



		 }
		 else  if(map[level][i-1][j]==BOX||map[level][i-1][j]==BOX+DEST)
		 {
			 if (map[level][i - 2][j] == SPACE || map[level][i - 2][j] == DEST)
			 {
				 map[level][i - 2][j] += BOX;
				 map[level][i - 1][j] = map[level][i - 1][j] - BOX + PLAYER;
				 map[level][i][j] -= PLAYER;//玩家消失在原来位置





			 }



		 }

		break;
	case 97:
	case 'a ':
	case'A':
		if (map[level][i][j-1] == SPACE || map[level][i][j-1] == DEST)
		{
			map[level][i][j-1] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i][j-1] == BOX || map[level][i][j-1] == BOX + DEST)
		{
			if (map[level][i][j-2] == SPACE || map[level][i][j-2] == DEST)
			{
				map[level][i][j-2] += BOX;
				map[level][i][j-1] = map[level][i][j-1] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 115:
	case 's ':
	case'S':
		if (map[level][i+1][j] == SPACE || map[level][i+1][j] == DEST)
		{
			map[level][i+1][j] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i+1][j] == BOX || map[level][i+1][j] == BOX + DEST)
		{
			if (map[level][i+2][j] == SPACE || map[level][i+2][j] == DEST)
			{
				map[level][i+2][j] += BOX;
				map[level][i+1][j] = map[level][i+1][j] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 100:
	case 'd ':
	case'D':
		if (map[level][i][j+1] == SPACE || map[level][i][j+1] == DEST)
		{
			map[level][i][j+1] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i][j+1] == BOX || map[level][i][j+1] == BOX + DEST)
		{
			if (map[level][i][j+2] == SPACE || map[level][i][j +2] == DEST)
			{
				map[level][i][j+2] += BOX;
				map[level][i][j+1] = map[level][i][j+1] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;








	}



}

bool jude()
{
	
	for (int i = 0; i < 10; i++)
	{
		for (int j = 0; j < 10; j++)
		{
			if (map[level][i][j] == BOX)
			{
				return false;
			}
		}
	}






}
int main()
{
	initgraph(10 * 40, 10 * 40);
	loadimg();
	system("mode con lines=15 cols=25");//调整窗口大小
	
	while (1)
	{
		//system("cls");
		gamedraw();
		if (jude())
		{
			level++;
			if (level > 2)
			{
				settextcolor(BLACK);
				settextstyle(25, 0, "微软雅黑");
				setbkmode(TRANSPARENT);
				
				outtextxy(100, 100, "oioioioioioioioi奥哈呦学妹你通过了！");
				_getch();
				break;
			}
		}
			keyevent();
			
		}
	getchar();//不让程序退出
	return 0;
		
	}

20.演示2

20231002_153153

21.关卡的增加

只需要将定义的全局变量map[3][10][10]，中3改成你想要的关卡数，然后在三维数组中增加像上面的二维数组即可，全部通关的level需要>关卡数-1即可

22.关卡重开

需要再定义一个和map[][][]三维数组相同的数组用来保存每一关的初始情况resetmap[][][],

c 复制代码 
int mapreset[3][10][10] =
{
	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,1,1,1,0,0,0,0},
	{0,0,0,1,2,1,1,1,1,0},
	{0,1,1,1,3,0,3,2,1,0},
	{0,1,2,3,4,0,1,1,1,0},
	{0,1,1,1,1,3,1,0,0,0},
	{0,0,0,0,1,2,1,0,0,0},
	{0,0,0,0,1,1,1,0,0,0},
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},

	{
	{0,0,0,0,0,0,0,0,0,0},
	{0,0,1,1,0,0,1,1,0,0},
	{0,1,0,2,1,1,2,0,1,0},
	{1,0,0,0,3,0,0,0,0,1},
	{1,0,0,0,4,3,0,0,0,1},
	{0,1,0,0,3,3,0,0,1,0},
	{0,0,1,0,0,0,0,1,0,0},
	{0,0,0,1,2,2,1,0,0,0},
	{0,0,0,0,1,1,0,0,0,0},
	{0,0,0,0,0,0,0,0,0,0}
	},
	{
	{0,0,0,0,1,0,0,0,0,0},
	{0,0,0,1,0,1,0,0,0,0},
	{0,0,1,2,3,0,1,0,0,0},
	{0,1,0,0,0,0,0,1,0,0},
	{1,2,3,0,4,0,0,0,1,0},
	{0,1,0,0,0,0,0,3,2,1},
	{0,0,1,0,3,0,0,0,1,0},
	{0,0,0,1,2,0,0,1,0,0},
	{0,0,0,0,1,0,1,0,0,0},
	{0,0,0,0,0,1,0,0,0,0}
	}
};

注意还是全局变量，我们设置当r键按下重置本关，所以我们要修改 keyevent()函数，当r按下我们遍历map数组将他赋值为原来的地图

c 复制代码 
void keyevent()
{
	int i = 0; int j = 0;
	for (i = 0; i < 10; i++)
	{
		for (j = 0; j < 10; j++)
		{
			if (map[level][i][j] == PLAYER || map[level][i][j] == PLAYER + DEST)
			{
				goto end;
			}
		}
	}
end:;
	char ch = _getch();
	//printf("%d    %c", ch, ch);//w 119 a 97 s 115  d 100
	switch (ch)
	{
	case 119:
	case 'w ':
	case 'W':
		if (map[level][i - 1][j] == SPACE || map[level][i - 1][j] == DEST)
		{
			map[level][i - 1][j] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i - 1][j] == BOX || map[level][i - 1][j] == BOX + DEST)
		{
			if (map[level][i - 2][j] == SPACE || map[level][i - 2][j] == DEST)
			{
				map[level][i - 2][j] += BOX;
				map[level][i - 1][j] = map[level][i - 1][j] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}

		break;
	case 97:
	case 'a ':
	case'A':
		if (map[level][i][j - 1] == SPACE || map[level][i][j - 1] == DEST)
		{
			map[level][i][j - 1] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i][j - 1] == BOX || map[level][i][j - 1] == BOX + DEST)
		{
			if (map[level][i][j - 2] == SPACE || map[level][i][j - 2] == DEST)
			{
				map[level][i][j - 2] += BOX;
				map[level][i][j - 1] = map[level][i][j - 1] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 115:
	case 's ':
	case'S':
		if (map[level][i + 1][j] == SPACE || map[level][i + 1][j] == DEST)
		{
			map[level][i + 1][j] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i + 1][j] == BOX || map[level][i + 1][j] == BOX + DEST)
		{
			if (map[level][i + 2][j] == SPACE || map[level][i + 2][j] == DEST)
			{
				map[level][i + 2][j] += BOX;
				map[level][i + 1][j] = map[level][i + 1][j] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 100:
	case 'd ':
	case'D':
		if (map[level][i][j + 1] == SPACE || map[level][i][j + 1] == DEST)
		{
			map[level][i][j + 1] += 4;
			map[level][i][j] -= 4;



		}
		else  if (map[level][i][j + 1] == BOX || map[level][i][j + 1] == BOX + DEST)
		{
			if (map[level][i][j + 2] == SPACE || map[level][i][j + 2] == DEST)
			{
				map[level][i][j + 2] += BOX;
				map[level][i][j + 1] = map[level][i][j + 1] - BOX + PLAYER;
				map[level][i][j] -= PLAYER;//玩家消失在原来位置





			}



		}
		break;
	case 'r':///新增
	case'R' :///新增
		for (int i = 0; i < 10; i++)///新增
		{///新增
			for (int j = 0; j < 10; j++)///新增
			{
				map[level][i][j] = mapreset[level][i][j];///新增




			}///新增
		}///新增
		break;///新增








	}///新增



}
相关推荐
敲敲敲-敲代码4 分钟前
游戏设计：推箱子【easyx图形界面/c语言】
c语言·开发语言·游戏
simple_ssn26 分钟前
【C语言刷力扣】1502.判断能否形成等差数列
c语言·算法·leetcode
ahadee1 小时前
蓝桥杯每日真题 - 第10天
c语言·vscode·算法·蓝桥杯
好想有猫猫1 小时前
【51单片机】LCD1602液晶显示屏
c语言·单片机·嵌入式硬件·51单片机·1024程序员节
军训猫猫头1 小时前
35.矩阵格式的一到一百数字 C语言
c语言·算法
Mr_Xuhhh2 小时前
递归搜索与回溯算法
c语言·开发语言·c++·算法·github
爱吃生蚝的于勒4 小时前
C语言内存函数
c语言·开发语言·数据结构·c++·学习·算法
失落的香蕉7 小时前
C语言串讲-2之指针和结构体
java·c语言·开发语言
ChoSeitaku10 小时前
链表循环及差集相关算法题|判断循环双链表是否对称|两循环单链表合并成循环链表|使双向循环链表有序|单循环链表改双向循环链表|两链表的差集(C)
c语言·算法·链表
DdddJMs__13510 小时前
C语言 | Leetcode C语言题解之第557题反转字符串中的单词III
c语言·leetcode·题解
热门推荐
01玄机平台应急响应—webshell查杀02Ubuntu 20.04使用Livox mid 360 测试 FAST_LIO03【HarmonyOS】HUAWEI DevEco Studio 下载地址汇总04新版微信小程序获取用户手机号05文件或文件夹名称中有空格如何批量去除06组基轨迹建模 GBTM的介绍与实现（Stata 或 R）07【TC3xx芯片】TC3xx芯片电源管理系统PMS详解08优化手机性能，解决卡顿问题：关闭这3个微信开关，释放内存空间09基于YOLOv10深度学习的CT扫描图像肾结石智能检测系统【python源码+Pyqt5界面+数据集+训练代码】深度学习实战、目标检测10【Flutter】基于 GetX 实现 Dio 的生命周期自动化